Oxidation states NEET 2026: Decreasing order explained

Oxidation states NEET 2026: which sequence actually decreases?

Option B is correct in this oxidation states NEET 2026 question: nitrogen, aluminium and sulfur have oxidation numbers that decrease from left to right.

N (+ 5) > A l (+ 3) > S (- 2)

The supplied source is NEET UG 2026, re-held on 2026-06-21, Code-50. “Medium” is the question bank’s difficulty tier, not an official NTA classification.

Select the sequence in which the oxidation numbers of the designated atoms decrease:

A) P_{4} O_{10} > S O_{3} > H_{2} O

B) N_{2} O_{5} > A l_{2} O_{3} > H_{2} S

C) P b O_{2} > N_{2} O_{3} > S O_{3}

D) P_{4} O_{6} > C l_{2} O_{7} > A l H_{3}

The supplied transcription has lost the underlining. We use only comparison atoms established by the worked solution; we do not assume which atom was underlined in water. A fails before that missing detail matters.

How do oxidation numbers balance a compound’s charge?

For every neutral compound here, the oxidation numbers multiplied by their respective atom counts must sum to zero. Follow the supplied solution’s NCERT-rule method: assign known values, include every subscript, then solve for the unknown.

\sum (atom count \times oxidation number) = 0

The unknown is the oxidation number of one target atom. Two such atoms contribute twice that value: $x = oxidation number of one target atom$ $Contribution from two target atoms = 2 x$

Use these assignments:

Oxygen in the listed oxides: negative two.
Hydrogen in hydrogen sulfide: positive one.
Hydrogen in a metal hydride such as aluminium hydride: negative one, not positive one.

A compound’s zero net charge does not mean that each element has oxidation number zero. Their weighted contributions cancel.

Why is option B correct?

Option B gives nitrogen at positive five, aluminium at positive three and sulfur at negative two. Both adjacent comparisons decrease. Charge balance gives each value as follows.

Nitrogen in dinitrogen pentoxide:

N_{2} O_{5} : 2 x + 5 (- 2) = 0 \Rightarrow 2 x - 10 = 0 \Rightarrow 2 x = 10 \Rightarrow x = + 5

Five oxygen atoms contribute negative ten in total. Two nitrogen atoms must balance that, so each nitrogen has oxidation number positive five.

Aluminium in aluminium oxide:

A l_{2} O_{3} : 2 x + 3 (- 2) = 0 \Rightarrow 2 x - 6 = 0 \Rightarrow 2 x = 6 \Rightarrow x = + 3

The positive six contribution belongs to two aluminium atoms, not one. Dividing by the aluminium atom count gives positive three.

Sulfur in hydrogen sulfide:

H_{2} S : 2 (+ 1) + x = 0 \Rightarrow 2 + x = 0 \Rightarrow x = - 2

Substitute the calculated values: $+ 5 > + 3 > - 2$

Decreasing order compares signed numbers, not absolute magnitudes. Keep sulfur’s minus sign: negative two lies below both positive values. The answer is option B.

Where do options A, C and D fail?

A and D fail at their first comparison; C fails at its second. One false adjacent inequality invalidates the whole decreasing sequence. Use the same charge-balance method to find where each option breaks.

A: phosphorus is below sulfur, not above it.

P_{4} O_{10} : 4 x + 10 (- 2) = 0 \Rightarrow 4 x = 20 \Rightarrow x = + 5

S O_{3} : x + 3 (- 2) = 0 \Rightarrow x = + 6

+ 5 > + 6 is false

A is eliminated before reaching water. Its missing underlining therefore does not affect the answer.

C: the first comparison works, but the second fails.

P b O_{2} : x + 2 (- 2) = 0 \Rightarrow x = + 4

N_{2} O_{3} : 2 x + 3 (- 2) = 0 \Rightarrow 2 x = 6 \Rightarrow x = + 3

Reuse the sulfur value already calculated:

+ 4 > + 3 > + 6 fails because + 3 > + 6 is false

D: chlorine is above phosphorus, not below it.

P_{4} O_{6} : 4 x + 6 (- 2) = 0 \Rightarrow 4 x = 12 \Rightarrow x = + 3

C l_{2} O_{7} : 2 x + 7 (- 2) = 0 \Rightarrow 2 x = 14 \Rightarrow x = + 7

+ 3 > + 7 is false

For completeness, hydrogen is negative one in aluminium hydride. This gives aluminium positive three, which cannot rescue the failed first comparison:

A l H_{3} : x + 3 (- 1) = 0 \Rightarrow x = + 3

Why does counting oxygen atoms incorrectly suggest A?

Ranking the formulas by oxygen count would incorrectly suggest A because they contain ten, three and one oxygen atoms respectively. That compares the wrong quantity: the question asks for an oxidation number per designated atom, not the number of oxygen atoms in a formula.

P_{4} O_{10}, S O_{3}, H_{2} O 10 > 3 > 1

In the phosphorus oxide, the oxygen contribution must be balanced across four phosphorus atoms. In sulfur trioxide, only one sulfur atom supplies the balancing contribution.

P: \frac{20}{4} = + 5 S: \frac{6}{1} = + 6

More oxygen atoms alone are not a universal shortcut across different compounds. The full formula and oxidation-number rules control the result.

Write the charge-balance equation before comparing oxidation numbers. Include the target atom’s subscript so you calculate the value for one atom, not its total contribution.

Can you apply the same method to three related questions?

The same charge-balance method handles reordered compounds, ions and hydrogen exceptions. These are original practice questions from Redox Reactions and Electrochemistry, not additional verified NEET PYQs. Attempt all three before reading the worked answers.

Practice prompts

Arrange the oxidation numbers of phosphorus, nitrogen and chlorine in the following compounds, respectively, in decreasing order:

P_{4} O_{6}, N_{2} O_{5}, C l_{2} O_{7}

Find sulfur’s oxidation number in the sulfate ion: $S O_{4}^{2 -}$
Find hydrogen’s oxidation number in each compound: $N a H, H_{2} O$

Worked answer 1

Apply oxygen’s negative-two value. Divide the balancing contribution by the target atom count:

P_{4} O_{6} : 4 x - 12 = 0 \Rightarrow x = + 3

N_{2} O_{5} : 2 x - 10 = 0 \Rightarrow x = + 5

C l_{2} O_{7} : 2 x - 14 = 0 \Rightarrow x = + 7

C l (+ 7) > N (+ 5) > P (+ 3)

Worked answer 2

For an ion, the sum equals the ion charge, not zero. Sulfur therefore has oxidation number positive six:

x + 4 (- 2) = - 2 \Rightarrow x - 8 = - 2 \Rightarrow x = + 6

Worked answer 3

Sodium has oxidation number positive one. Hydrogen must therefore be negative one in sodium hydride:

N a H : + 1 + x = 0 \Rightarrow x = - 1

In water, oxygen is negative two, so hydrogen is positive one. Here hydrogen is explicitly the practice question’s target, unlike the lost underlining in option A.

H_{2} O : 2 x - 2 = 0 \Rightarrow x = + 1

Identify the target atom, assign known values, balance the charge, then compare signed results.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

Keep going with Standing Waves in Pipes NEET 2026: Node-Count Ratio.

Frequently asked questions

Which option has decreasing oxidation numbers in this NEET 2026 question?

Option B is correct: nitrogen in N2O5 is +5, aluminium in Al2O3 is +3 and sulfur in H2S is -2. The sequence +5 > +3 > -2 decreases at both adjacent comparisons.

How do I calculate oxidation numbers using charge balance?

Multiply each element's oxidation number by its atom count and add the contributions. The sum equals zero for a neutral compound and the net charge for an ion. For nitrogen in N2O5, 2x + 5(-2) = 0 gives x = +5.

Why can't I rank oxidation numbers by counting oxygen atoms?

Oxygen count alone ignores how many target atoms share the balancing contribution. In P4O10, each phosphorus is +5, while sulfur in SO3 is +6 despite the lower oxygen count. This makes option A's first comparison, +5 > +6, false.

Is hydrogen's oxidation number always +1?

No. Hydrogen is +1 in H2S and H2O, but -1 in metal hydrides such as NaH and AlH3. Using hydrogen as -1 in AlH3 gives aluminium an oxidation number of +3.